Given a cofiltered diagram of commutative rings $F:D\to \mathrm{CRing}$, we obtain a filtered diagram $\mathrm{Mod}(F):D^{op}\to \mathrm{ExAbCat}$ (where $\mathrm{ExAbCat}$ is the category of Abelian categories with exact additive functors between them) induced by the contravariant functor $CRing^{op}\to \mathrm{ExAbCat}$ sending the maps $f:A\to B$ in $\mathrm{CRing}$ to the restriction functor, $f^*:B\mathrm{-Mod}\to A\mathrm{-Mod}$.

Then here's the question:
Is $\varinjlim \mathrm{Mod}(F)$ equivalent to $(\varprojlim F)\mathrm{-Mod}$?

Do you really mean to ask whether limits in the category of commutative rings are taken to colimits in the (2-)category of abelian categories? Or did you actually mean to ask whether (filtered) colimits in commutative rings are taken to limits in the category of abelian categories? I believe the answer is "no" for the first and "yes" for the second (under reasonable interpretations, which depend on the exact interpretation of the category or 2-category of abelian categories), and I think I'm prepared to say more, but I need to ask first whether you're sure this is what you meant to ask.
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Todd Trimble♦Apr 9 '11 at 11:22

Thank you for your quick response! I actually meant the first. If it was not entirely clear what I meant, it is probably because I am still a bit confused :-) So, the answer is "no" I guess. Is there an easy way to see that?
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JvWApr 9 '11 at 12:21

Oh, and if there are still details you need to know, which I was unable to express, let me know, I will try take make myself clearer.
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JvWApr 9 '11 at 12:40

Except, Harry, I think you got the arrows under the limits reversed in your edit: the backwards-pointing arrow is a limit and the forwards-pointing one is the colimit, which is not the way the original question had it. I will change it for you.
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Todd Trimble♦Apr 9 '11 at 15:02

1 Answer
1

Unless I am severely mistaken, the answer to the question as stated is "no". (There is by the way some slight disagreement in the literature as to what "filtered category" means; for many, a filtered poset means a nonempty poset where any two elements have an upper bound, but in Sheaves in Geometry and Logic, Mac Lane and Moerdijk mean that any two elements have a lower bound. I don't think this affects my answer.)

Let us take for instance the ring of $p$-adic integers as inverse limit of the (directed in either sense) system

$$\ldots \to \mathbb{Z}/(p^{n+1}) \to \mathbb{Z}/(p^n) \to \ldots$$

Each of these commutative rings $A$ can be viewed as an $Ab$-enriched category, and the abelian category of left modules over $A$ can be viewed as the $Ab$-enriched category of $Ab$-enriched functors $A \to Ab$. So we are homming into $Ab$, and the usual expectation is that a contravariant hom-functor takes colimits to limits, but not limits to colimits; the question as stated has to do with the latter situation.

The filtered colimit (and it would be reasonable to interpret "colimit" in a 2-categorical sense) of the system of full subcategory inclusions

can be taken to be the union of these (abelian) categories. The union is just the category of those abelian groups where each object is completely annihilated by some $p^n$ (not uniformly of course; the $n$ depends on the group); in particular, they are all torsion. On the other hand, most objects in the category of modules over the $p$-adic integers, for example the $p$-adic integers itself, are not torsion. So there's your counterexample.

But if we turn the directions around, then indeed colimits are taken to limits. There are some slightly tricky issues to deal with, because "morally" we really ought to be working with 2-categorical limits -- weak limits to be precise. To do it right, we should be working with the category of commutative rings as a 2-category (construing ring homomorphisms as enriched functors and taking 2-cells as natural transformations), but if I'm not mistaken, weak colimits in this 2-category of commutative rings coincide with ordinary colimits of commutative rings, and we can relax. Anyway, it is a general feature that homming into an object like $Ab$, giving here a functor

Anyway, I think that it's actually strictly true by merit of the fact that the Mod(-) functor is a left-adjoint. Specifically, it's left-adjoint to the functor sending an abelian category to the ring of endomorphisms of the identity functor.
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Harry GindiApr 9 '11 at 15:11

Harry, that doesn't seem right to me. If for example $A = Ab$, then you are saying that exact additive functors $Mod(R) \to Ab$ are somehow equivalent to ring homomorphisms $R \to \mathbb{Z}$?
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Todd Trimble♦Apr 9 '11 at 15:26